A compound consists of 47.8% zinc and 52.2% chlorine by mass. The empirical formula is `Zn_x Cl_y` where x and y can have the values

Options

A. 1 and 1
B. 1 and 2
C. 2 and 1
D. 2 and 3 respectively

Explanation

To determine the empirical formula of the compound consisting of 47.8% zinc and 52.2% chlorine by mass, we will follow these steps:

### Step 1: Calculate the number of moles of Zinc
1. Given mass of Zinc = 47.8 g
2. Molar mass of Zinc (Zn) = 65.38 g/mol (approximately)
3. Number of moles of Zinc = Mass / Molar mass = 47.8 g / 65.38 g/mol

\[ \text{Moles of Zn} = \frac{47.8}{65.38} \approx 0.731 \text{ moles} \]

### Step 2: Calculate the number of moles of Chlorine
1. Given mass of Chlorine = 52.2 g
2. Molar mass of Chlorine (Cl) = 35.45 g/mol (approximately)
3. Number of moles of Chlorine = Mass / Molar mass = 52.2 g / 35.45 g/mol

\[ \text{Moles of Cl} = \frac{52.2}{35.45} \approx 1.474 \text{ moles} \]

### Step 3: Determine the simplest mole ratio
1. We have approximately 0.731 moles of Zinc and 1.474 moles of Chlorine.
2. To find the simplest ratio, divide each value by the smallest number of moles (which is 0.731):

\[ \text{Ratio of Zn} = \frac{0.731}{0.731} = 1 \]
\[ \text{Ratio of Cl} = \frac{1.474}{0.731} \approx 2.017 \approx 2 \]

### Step 4: Write the empirical formula
From the ratios calculated:
- The ratio of Zinc (Zn) is 1
- The ratio of Chlorine (Cl) is 2

Thus, the empirical formula is:
\[ \text{Empirical formula} = \text{Zn}_1\text{Cl}_2 \text{ or simply } \text{ZnCl}_2 \]

### Conclusion
The values of x and y in the empirical formula Zn_x Cl_y are:
- x = 1
- y = 2